All categories

3 years ago

What is an equation in slope-intercept form of the line that passes through the points (−2, −2) and (1, 7)?

Mathematics

2 answers:

Setler79 [48]3 years ago

7 0

Your answer could be b or d but I think it is b using the equation y=mx +b

Nana76 [90]3 years ago

5 0

B) since you can use rise over run? 9/3 is 3 so the slope is 3x, but then you plug in one of the points to find the y intercept. I would use (1,7) for simplicity. It would be B

You might be interested in

A store owner is interested in opening a second shop. She wants to estimate the true average daily revenue of her current shop t

Alex

Every confidence interval has associated z value. As confidence interval increases so do the z value associated with it.
The confidence interval can be calculated using following formula:
$\overline{x} \pm \frac{zs}{\sqrt n}$
Where $\overline{x}$ is the mean value, z is the associated z value, s is the standard deviation and n is the number of samples.
We know that standard deviation is simply a square root of variance:
$s=\sqrt{315900.20}=\$562.05$
The confidence interval of 95% has associated z value of <span>1.960.
</span>Now we can calculate the confidence interval for our income:
$3472.00\pm \frac{1.960\cdot 562.05}{\sqrt{60}}\\ \$3472.00\pm142.22$

6 0

4 years ago

What is 3/4 h = – 45/52

sergiy2304 [10]

Mixed number form: h = - 1 2/13 Hope that helps :))

6 0

3 years ago

mei has 15 oranges,9 peaches, and 18 pears. she wants to put all of the fruit into decorative baskets, what's the greats number

pashok25 [27]

Answer:

15,9,8 can be divided by 3

so, the greatest number of fruit in each basket is 3 Step-by-step explanation:

6 0

3 years ago

Find each missing measure.

oksano4ka [1.4K]

The answer

a = 84

ac = 9

6 0

3 years ago

Each year for 4 years, a farmer increased the number of trees in a certain orchard by of the number of trees in the orchard the

Neko [114]

Answer:

The number of trees at the begging of the 4-year period was 2560.

Step-by-step explanation:

Let’s say that x is number of trees at the begging of the first year, we know that for four years the number of trees were incised by 1/4 of the number of trees of the preceding year, so at the end of the first year the number of trees was $x+\frac{1}{4} x=\frac{5}{4} x$ , and for the next three years we have that

Start End

Second year $\frac{5}{4}x$ -------------- $\frac{5}{4}x+\frac{1}{4}(\frac{5}{4}x) =\frac{5}{4}x+ \frac{5}{16}x=\frac{25}{16}x=(\frac{5}{4} )^{2}x$

Third year $(\frac{5}{4} )^{2}x$ ------------- $(\frac{5}{4})^{2}x+\frac{1}{4}((\frac{5}{4})^{2}x) =(\frac{5}{4})^{2}x+\frac{5^{2} }{4^{3} } x=(\frac{5}{4})^{3}x$

Fourth year $(\frac{5}{4})^{3}x$ -------------- $(\frac{5}{4})^{3}x+\frac{1}{4}((\frac{5}{4})^{3}x) =(\frac{5}{4})^{3}x+\frac{5^{3} }{4^{4} } x=(\frac{5}{4})^{4}x.$

So the formula to calculate the number of trees in the fourth year is

$(\frac{5}{4} )^{4} x,$ we know that all of the trees thrived and there were 6250 at the end of 4 year period, then

$6250=(\frac{5}{4} )^{4}x$ ⇒ $x=\frac{6250*4^{4} }{5^{4} }= \frac{10*5^{4}*4^{4} }{5^{4} }=2560.$

Therefore the number of trees at the begging of the 4-year period was 2560.

7 0

3 years ago